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Alex787 [66]
3 years ago
15

Unlike most real bulbs, the resistances of the bulbs in the questions below do not change as the current through them changes. A

ll bulbs considered in this problem are identical. Assume all batteries are ideal (they have no internal resistance) and connecting wires have no resistance.
1. The current passing through the brighter

bulb is larger.

2. Not enough information is given.

3. The current passing through the brighter

bulb is smaller.

4. The current passing through both bulbs

are the same.
Physics
1 answer:
Shkiper50 [21]3 years ago
5 0

Answer:

Option 1 is correct.

The current passing through the brighter bulb is larger.

Explanation:

The brightness of the bulb is determined by the power, I²R

And since they all have equal resistances, the only factor different that could result in more or less power is the current, I through the bulb.

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Which is not a factor that affects the pressure of a gas in a closed container?
maksim [4K]

it is size of of particles because it does not matter about the size in a closed container

7 0
3 years ago
What is the order of magnitude of the distance of Sun to nearest star in meters?
neonofarm [45]

Answer:

Approximating the Milky Way as a disk and using the density in the solar neighborhood, there are about 100 billion stars in the Milky Way.

Explanation:

Since we are making an order of magnitude estimate, we will make a series of simplifying assumptions to get an answer that is roughly right.

Let's model the Milky Way galaxy as a disk.

The volume of a disk is:

V

=

π

⋅

r

2

⋅

h

Plugging in our numbers (and assuming that

π

≈

3

)

V

=

π

⋅

(

10

21

m

)

2

⋅

(

10

19

m

)

V

=

3

×

10

61

m

3

Is the approximate volume of the Milky Way.

Now, all we need to do is find how many stars per cubic meter (

ρ

) are in the Milky Way and we can find the total number of stars.

Let's look at the neighborhood around the Sun. We know that in a sphere with a radius of

4

×

10

16

m there is exactly one star (the Sun), after that you hit other stars. We can use that to estimate a rough density for the Milky Way.

ρ

=

n

V

Using the volume of a sphere

V

=

4

3

π

r

3

ρ

=

1

4

3

π

(

4

×

10

16

m

)

3

ρ

=

1

256

10

−

48

stars /

m

3

Going back to the density equation:

ρ

=

n

V

n

=

ρ

V

Plugging in the density of the solar neighborhood and the volume of the Milky Way:

n

=

(

1

256

10

−

48

m

−

3

)

⋅

(

3

×

10

61

m

3

)

n

=

3

256

10

13

n

=

1

×

10

11

stars (or 100 billion stars)

Is this reasonable? Other estimates say that there are are 100-400 billion stars in the Milky Way. This is exactly what we found.

4 0
2 years ago
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A train accelerates at -1.5 m/s2 for 10 seconds. If the train had an initial
Gnom [1K]

Answer:

17 m/s

Explanation:

Using formula a = (v-u) /t

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final velocity v = unknown

initial velocity u = 32 m/s

time t = 10s

-1.5 = (v-32)/10

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7 0
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Answer:

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Explanation:

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102=1/2 m 5²

m=8.16 Kg

8 0
2 years ago
What is the sharpness of the light and the amount of voltage when you put the magnet through the lower number of coils first vs
fomenos

Answer:hjvuihi

Explanation:lk

4 0
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